Question

Eight equal drops of water are falling through air with a steady speed of \( 10 \, \text{cm/s} \). If the drops coalesce, the new velocity is:

• A. \( 10 \, \text{cm/s} \)
• B. \( 40 \, \text{cm/s} \)
• C. \( 16 \, \text{cm/s} \)
• D. \( 5 \, \text{cm/s} \)

Hint:

When drops merge, the radius of the combined drop increases as the cube root of the total volume. Since terminal velocity depends on the square of the radius, the new velocity scales quadratically with the radius increase.

Answer 1

The Correct Option is B

When the drops coalesce, their total volume remains unchanged. Let: - \( r \) be the radius of each small drop, - \( R \) be the radius of the combined drop. 

Step 1: Apply Volume Conservation The total volume of eight small drops is: \[ 8 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3. \] Simplify: \[ R^3 = 8r^3 \quad \implies \quad R = 2r. \] 

Step 2: Terminal Velocity Relation The terminal velocity \( v_T \) of a drop is proportional to the square of its radius: \[ v_T \propto r^2. \] Let \( v_1 = 10 \, \text{cm/s} \) be the terminal velocity of the small drops, and \( v_2 \) be the terminal velocity of the combined drop. Then: \[ \frac{v_1}{v_2} = \left( \frac{r}{R} \right)^2. \] Substitute \( R = 2r \): \[ \frac{v_1}{v_2} = \left( \frac{r}{2r} \right)^2 = \frac{1}{4}. \] 

Step 3: Calculate \( v_2 \) Rearrange to find \( v_2 \): \[ v_2 = 4v_1 = 4 \cdot 10 = 40 \, \text{cm/s}. \] 

Final Answer: \[ \boxed{40 \, \text{cm/s}} \]